Question

A 10.00-mL sample of vinegar, an aqueous solution of acetic acid $\left({\mathbf{HC}}_2{\mathbf{H}}_3{\mathbf{O}}_2\right)$, is titrated with 0.5062 M $\mathbf{NaOH}$, and 16.58 mL is required to reach the endpoint.

a. What is the molarity of the acetic acid?

b. If the density of the vinegar is$\mathbf{1}\mathbf{.}\mathbf{006}\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{${\mathbf{cm}}^3$}\right.$ . What is the mass percent of acetic acid in the vinegar?

Short answer

a. The molarity of acetic acid is 0.839M.

b.The mass percent of acetic acid in the vinegar is 5.01%.

Step-by-step solution

Definition of acetic acid

Acetic acid is a by-product of the fermentation process. It gives vinegar its characteristic odor. Vinegar has about 4-6% acetic acid in water.

Calculation of molarity of acetic acid

Here, $n\left(H{C}_2{H}_3{O}_2\right)=n\left(NaOH\right)$.

$c_1\times V_1=c_2\times V_2$

Now,

$c\left(H{C}_2{H}_3{O}_2\right)=\frac{c\left(NaOH\right)\times V\left(NaOH\right)}{V\left(H{C}_2{H}_3{O}_2\right)}$

$\begin{array}{l}c\left(H{C}_2{H}_3{O}_2\right)=\frac{0.5062M\times 16.58mL}{10mL}\\ c\left(H{C}_2{H}_3{O}_2\right)=0.839\end{array}$

Hence,the molarity is 0.839.

Calculation of mass percentage

$\begin{array}{l}m\left(H{C}_2{H}_3{O}_2\right)=V\times c\times M_r\\ m\left(H{C}_2{H}_3{O}_2\right)=0.01658mL\times 0.5062M\times 60\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\\ m\left(H{C}_2{H}_3{O}_2\right)=0.5036g\end{array}$

Now,

$\begin{array}{l}m\left(vinegar\right)=ro\times V\\ m\left(vinegar\right)=1.006\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$c{m}^3$}\right.\times 10mL\\ m\left(vinegar\right)=10.069g\end{array}$

The mass percentage is,

$\begin{array}{l}m\%=\frac{m\left(H{C}_2{H}_3{O}_2\right)}{m\left(vinegar\right)}\times 100\\ m\%=\frac{0.5036g}{10.069g}\times 100\\ m\%=5.01\%\end{array}$

Hence, the mass percentage is 5.01%.